Delta-Function Potential

The attractive delta-function potential is the canonical one-dimensional singular well. It is an idealized zero-range interaction whose entire effect is encoded in a discontinuity of the derivative of the wavefunction.

The standard attractive model is

$$V(x)=-\alpha\delta(x), \qquad \alpha\gt0.$$

The parameter $\alpha$ has units of energy times length. Physically, this potential is the limit of a narrow attractive well whose width goes to zero while its area stays fixed.

Hamiltonian And Matching Condition

The stationary Schrödinger equation is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} -\alpha\delta(x)\psi(x) =E\psi(x).$$

For $x\ne 0$, the particle is free. The singular point at the origin supplies the matching condition. Integrate the equation from $-\epsilon$ to $+\epsilon$:

$$-\frac{\hbar^2}{2m} \left[ \psi'(\epsilon)-\psi'(-\epsilon) \right] -\alpha\psi(0) = E\int_{-\epsilon}^{\epsilon}\psi(x)\,dx.$$

If $\psi$ is finite at the origin, the integral on the right vanishes as $\epsilon\to 0$. Thus

$$\psi'(0^+)-\psi'(0^-) =-\frac{2m\alpha}{\hbar^2}\psi(0).$$

The wavefunction itself remains continuous:

$$\psi(0^-)=\psi(0^+)=\psi(0).$$

This pair of conditions replaces ordinary derivative continuity. Requiring $\psi'$ to be continuous would remove the delta interaction.

Bound-State Solution

A bound state has $E\lt0$. Define

$$\kappa=\frac{\sqrt{-2mE}}{\hbar}, \qquad \kappa\gt0.$$

For $x\ne 0$, the equation becomes

$$\frac{d^2\psi}{dx^2}=\kappa^2\psi.$$

Square integrability requires exponential decay away from the origin, so the bound-state wavefunction has the form

$$\psi(x)=A e^{-\kappa\lvert x\rvert}.$$

Its derivative is

$$\psi'(0^+)=-\kappa A, \qquad \psi'(0^-)=+\kappa A.$$

Therefore

$$\psi'(0^+)-\psi'(0^-) =-2\kappa A.$$

The jump condition gives

$$-2\kappa A =-\frac{2m\alpha}{\hbar^2}A,$$

so

$$\kappa=\frac{m\alpha}{\hbar^2}.$$

The energy is

$$E =-\frac{\hbar^2\kappa^2}{2m} =-\frac{m\alpha^2}{2\hbar^2}.$$

There is exactly one bound state for the attractive delta potential.

Normalization

Normalize the state by imposing

$$1 =\int_{-\infty}^{\infty} \lvert \psi(x)\rvert^2\,dx.$$

For $\psi(x)=A e^{-\kappa\lvert x\rvert}$,

$$1 =2\lvert A\rvert^2 \int_0^\infty e^{-2\kappa x}\,dx =\frac{\lvert A\rvert^2}{\kappa}.$$

Choosing $A$ real and positive gives

$$A=\sqrt{\kappa}.$$

The normalized bound state is therefore

$$\psi(x) =\sqrt{\kappa}\,e^{-\kappa\lvert x\rvert}, \qquad \kappa=\frac{m\alpha}{\hbar^2}.$$

The characteristic localization length is $1/\kappa=\hbar^2/(m\alpha)$. A stronger attraction produces a narrower, more deeply bound state.

Relation To A Narrow Square Well

The delta potential can be obtained as a limiting model. Consider an attractive square well of width $\ell$ and depth $V_0$ centered at the origin, with

$$V_0\ell=\alpha.$$

Take

$$\ell\to 0, \qquad V_0\to\infty, \qquad V_0\ell=\alpha \quad\text{fixed}.$$

In this limit the detailed shape of the well disappears, but its integrated strength remains. The wavefunction cannot resolve the interior structure of the narrow well; it only feels the matching condition at the origin.

This is why the delta potential is useful as an effective zero-range model. It captures a low-resolution interaction without pretending to describe the microscopic force profile.

Repulsive Delta Potential

For a repulsive delta potential,

$$V(x)=+\alpha\delta(x), \qquad \alpha\gt0,$$

the jump condition becomes

$$\psi'(0^+)-\psi'(0^-) =+\frac{2m\alpha}{\hbar^2}\psi(0).$$

A normalizable bound state would still have the form $A e^{-\kappa\lvert x\rvert}$, whose derivative jump is negative when $A\ne 0$. It cannot satisfy the positive jump condition. Therefore a repulsive delta potential has no bound state.

Scattering Preview

The delta potential also has a simple scattering problem. For a general singular interaction

$$V(x)=\lambda\delta(x),$$

the matching condition is

$$\psi'(0^+)-\psi'(0^-) =\frac{2m\lambda}{\hbar^2}\psi(0).$$

For an incoming plane wave from the left,

$$\psi(x)= \begin{cases} e^{ikx}+r e^{-ikx}, & x\lt0,\\ t e^{ikx}, & x\gt0, \end{cases}$$

continuity and the derivative jump determine the reflection and transmission amplitudes. This is the simplest example where a potential localized at a single point changes scattering even though the particle is free everywhere else.

The full elementary scattering calculation is Scattering from a Delta Potential. Its probability-current interpretation uses the same ideas as Potential Step and Rectangular Barrier Tunneling. Delta Potential Scattering then verifies that this bound state reappears as the attractive even-channel pole and relates the pole residue to the normalized exponential tail.

Physical Lessons

The delta well teaches several durable lessons:

• singular potentials are defined by matching conditions, not by ordinary pointwise force intuition;
• finite jumps in $V(x)$ preserve derivative continuity, but delta functions do not;
• a one-dimensional attractive zero-range potential supports one bound state;
• the bound-state size is controlled by the inverse integrated strength;
• scattering can be affected by an interaction concentrated at a single point.

The model is simple enough for exact calculation but subtle enough to expose why domains and boundary conditions matter. The operator-domain warning is developed more generally in Hermitian vs Self-Adjoint Operators.

Common Mistakes

• Requiring $\psi'$ to be continuous across the delta potential.
• Forgetting that $\psi$ itself remains continuous for this standard delta interaction.
• Treating $\delta(x)\psi(x)$ as an ordinary product of functions instead of using the integrated matching condition.
• Using the attractive bound-state formula for the repulsive delta potential.
• Forgetting that $\alpha$ has units of energy times length.
• Confusing the zero width of the idealized potential with zero physical effect.

Where This Is Used

• Boundary Conditions derives the general jump-condition rule for delta potentials.
• Delta Function gives the distributional background.
• Finite Square Well is the finite-width model whose narrow limit motivates the delta well.
• Double Delta Potential uses the same jump condition twice to produce bonding and antibonding bound states.
• Scattering from a Delta Potential solves the corresponding point-interaction scattering problem.
• Delta Potential Scattering connects that scattering solution to its parity-channel pole and residue.
• Energy Scales in One Dimension highlights the delta-well binding length and energy scale.
• Distributions explains why delta functions must be handled under integrals.
• Later tunneling and scattering pages reuse the same continuity-plus-jump method.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• S. Flügge, Practical Quantum Mechanics, Springer, 1999.

Exercises

1. Derive the derivative jump condition for $V(x)=\lambda\delta(x)$.

Solution

Start from

$$-\frac{\hbar^2}{2m}\psi''(x) +\lambda\delta(x)\psi(x) =E\psi(x).$$

Integrate from $-\epsilon$ to $+\epsilon$:

$$-\frac{\hbar^2}{2m} \left[ \psi'(\epsilon)-\psi'(-\epsilon) \right] +\lambda\psi(0) = E\int_{-\epsilon}^{\epsilon}\psi(x)\,dx.$$

Taking $\epsilon\to 0$ makes the right-hand side vanish for finite $\psi$. Therefore

$$-\frac{\hbar^2}{2m} \left[ \psi'(0^+)-\psi'(0^-) \right] +\lambda\psi(0) =0,$$

or

$$\psi'(0^+)-\psi'(0^-) =\frac{2m\lambda}{\hbar^2}\psi(0).$$

2. Normalize $\psi(x)=A e^{-\kappa\lvert x\rvert}$.

Solution

Compute

$$\int_{-\infty}^{\infty} \lvert A\rvert^2 e^{-2\kappa\lvert x\rvert}\,dx =2\lvert A\rvert^2 \int_0^\infty e^{-2\kappa x}\,dx =\frac{\lvert A\rvert^2}{\kappa}.$$

Setting this equal to $1$ gives $\lvert A\rvert=\sqrt{\kappa}$. With a positive real phase convention,

$$A=\sqrt{\kappa}.$$

3. Show that the attractive delta potential has no odd bound state.

Solution

For $x\ne 0$, a negative-energy bound state must decay exponentially on both sides. An odd candidate would have opposite signs on the two sides. But the standard delta potential requires the wavefunction to be continuous at the origin:

$$\psi(0^-)=\psi(0^+).$$

An odd continuous wavefunction has $\psi(0)=0$. The jump condition then gives

$$\psi'(0^+)-\psi'(0^-)=0.$$

The only exponentially decaying odd solution compatible with both continuity and derivative continuity at the origin is the zero function. Therefore the attractive delta well has only the even bound state.